求不定积分 √(x^2-a^2)dx/x^2 求过程 答案是 ln|x+√(x^2-a^2)|...
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发布时间：2024-04-16 04:23
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热心网友
时间：2024-04-18 23:38
=∫√((x/a)^2-1)d(x/a)/(x/a)^2
x/a=secu cosu=a/x sinu=√(x^2-a^2)/x
d(x/a)=secutanudu
=∫tanu*secu*tanudu/(secu)^2
=∫(secu^3-secu)du/(secu)^2
=∫secudu-∫cosudu
=∫du/cosu -sinu
=∫dsinu/[(1-sinu)(1+sinu)] -sinu
=(1/2)∫d(1+sinu)/(1+sinu)-(1/2)∫d(1-sinu)/(1-sinu) -sinu
=ln|(1+sinu)/cosu| -sinu+C0
(1+sinu)/cosu=[x+√(x^2-a^2)]/a
=ln|x+√(x^2-a^2)|-ln|a| -√(x^2-a^2)/x+C0
=ln|x+√(x^2-a^2)| -√(x^2-a^2)/x+C C=C0-lna
